Chapter 8, Problem 89QAP

To determine

(a)

Acceleration and the tension of the string

Answer to Problem 89QAP

Acceleration = $8.0{\text{ ms}}^{\text{-1}}$

Tension of the string = $32.0\text{ N}$

Explanation of Solution

Given info:

Mass of m₁= $2\text{ kg}$

Mass of m₂= $\text{4 kg}$

Radius of pulley = $0.04\text{ m}$

Mass of pulley = $0.5\text{ kg}$

Acceleration of the mass= $0.35{\text{ ms}}^{\text{-2}}$

Formula used:

  $\begin{array}{l}F=ma\\ F=\text{ Force}\\ m=\text{ mass of pulley}\\ a=\text{ accleration}\end{array}$

Calculation:

Considering the forces on string,

  $\begin{array}{l}{\text{For the m}}_{\text{2}}\text{ vertically,}\\ m_{2}g-T=m_{2}a\\ m_2={\text{mass of m}}_{\text{2}}\\ T=\text{tension of the string(as it is connected to two masses, it is uniform)}\\ m_{2}g-T=m_{2}a\\ T=m_{2}g-m_{2}a\\ {\text{For the m}}_{\text{1}}\text{ there are no opposing forces to }T\text{ in horizontal direction}\\ F=ma\\ T=m_{1}a\end{array}$

  $\begin{array}{l}\text{As they are connected, }a\text{(acceleration) have to be equal, T is uniform}\\ \text{Combining two equations together,}\\ T=m_{2}g-m_{2}a\\ T=m_{1}a\\ m_{2}g-m_{2}a=m_{1}a\\ 4.0*10-4.0*a=2*4\\ a=8{\text{ ms}}^{\text{-2}}\\ T=m_{1}a\\ T=4.0*8\\ T=32.0\text{ N}\end{array}$

Conclusion:

Acceleration = $8.0{\text{ ms}}^{\text{-1}}$

Tension of the string = $32.0\text{ N}$

To determine

(b)

Time takes to travel $2.25\text{ m}$

Answer to Problem 89QAP

Time takes to travel $2.25\text{ m}$= $0.75\text{ s}$

Explanation of Solution

Given info:

Mass of m₁= $2\text{ kg}$

Mass of m₂= $\text{4 kg}$

Radius of pulley = $0.04\text{ m}$

Mass of pulley = $0.5\text{ kg}$

Traveled distance = $2.25\text{ m}$

Formula used:

  $\begin{array}{l}S=ut+\frac{1}{2}a{t}^2\\ S=\text{ distance}\\ u=\text{ initial velocity}\\ a=\text{ accleration}\\ t=\text{ time}\end{array}$

Calculation:

Assume, blocks start movement from the rest

  $\begin{array}{l}S=ut+\frac{1}{2}a{t}^2\\ S=ut+\frac{1}{2}a{t}^2\\ 2.25\text{ m=0*t+}\frac{1}{2}*8.0{\text{ ms}}^{\text{-2}}*t^2\\ t=0.75\text{ s}\end{array}$

Conclusion:

Time takes to travel $2.25\text{ m}$= $0.75\text{ s}$

To determine

(c)

Angular speed of the pulley

Answer to Problem 89QAP

Angular velocity/speed of the pulley = $150{\text{ rads}}^{\text{-1}}$

Explanation of Solution

Given info:

Mass of m₁= $2\text{ kg}$

Mass of m₂= $\text{4 kg}$

Radius of pulley = $0.04\text{ m}$

Mass of pulley = $0.5\text{ kg}$

Traveled distance = $2.25\text{ m}$

Formula used:

  $\begin{array}{l}v=u+at\\ v=\text{ final velocity}\\ u=\text{ initial velocity}\\ a=\text{ accleration}\\ t=\text{ time}\\ \text{v=r}\omega \\ v=\text{ translational velocity}\\ r=\text{ radius of the pulley}\\ \omega =\text{ angular velocity}\\ \text{6}{\text{.0 ms}}^{\text{-1}}\text{=0}{\text{.04 ms}}^{\text{-1}}*\omega \\ \omega =150{\text{ rads}}^{\text{-1}}\end{array}$

Calculation:

Assume, blocks start movement from the rest

  $\begin{array}{l}v=u+at\\ v=0+8.0{\text{ ms}}^{\text{-2}}*0.75\text{ s}\\ v=6.0{\text{ ms}}^{\text{-2}}\\ \text{Assume that the pulley rotates at the same linear velocity as the string}\\ \text{v=r}\omega \\ \text{6}{\text{.0 ms}}^{\text{-1}}\text{=0}{\text{.04 ms}}^{\text{-1}}*\omega \\ \omega =150{\text{ rads}}^{\text{-1}}\end{array}$

Conclusion:

Angular velocity/speed of the pulley = $150{\text{ rads}}^{\text{-1}}$